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Current time:0:00Total duration:6:53

AP.CALC:

FUN‑6 (EU)

, FUN‑6.B (LO)

, FUN‑6.C (LO)

so we have f of X being equal to the absolute value of x plus 2 and when we will and we want to evaluate the definite integral from negative 4 to 0 of f of X DX and like always pause this video and see if you could work through this now when you first do this you might stumble around a little bit because how do you take the antiderivative of an absolute value function and the key here is to one way to approach it is to rewrite f of X without the absolute value and we can do that by rewriting it as a piecewise function and the way I'm going to do it I'm going to think about intervals where what we what we where whatever we take inside the absolute value is going to be positive and other intervals where everything that we take inside the absolute value is going to be negative and the the point at which we change is where X plus 2 is equal to 0 or X is equal to negative 2 so let's just think about the intervals X is less than negative 2 and X is greater than or equal to negative 2 and this could have been less than or equal in which case this would have been greater than either way it would have been equal to this absolute value this is a continuous function here and so when well let's do the easier case when X is greater than or equal to negative 2 then X plus 2 is going to be positive or it's going to be it's going to be greater than or equal to zero and so the absolute value of it it's just going to be the X plus 2 so it's going to be X plus 2 when X is greater than or equal to negative 2 and what about when X is less than negative 2 well when X is less than negative 2 X plus 2 is going to be negative and then if you take the absolute value of a negative number you're going take the opposite of it so this is going to be negative x plus 2 and to really help grok this because frankly this is the hardest part of what we're doing and really this is more algebra than calculus let me draw the absolute value function to make this clear so that is my x axis that is my y axis and let's say we're here at negative 2 and so when we are less than negative 2 when X is less than negative 2 my graph is going to look like this it is going to it is going to look look something let's see what is it's going to look like that and when we are greater than negative two we do that in a different color when we are greater than negative two it's going to look like this it's going to look like that and so notice this is in blue we have this is the graph X plus two we could say this is a graph of y equals x plus 2 and what we have in magenta right over here this is the graph of negative X minus 2 it has a negative slope and we intercept the y-axis at negative 2 so it makes sense there's multiple ways that you could reason through this now once we break it up then we can break up the integral we could say that what we wrote here this is equal to the integral from negative 4 to 2 sorry negative 4 to negative 2 of f of X which is which is in that case it's going to be negative X minus 2 I just distributed the negative sign there D X and then plus plus the definite integral going from negative 2 to 0 of X plus 2 DX and just to make sure we know what we're doing here this if this is negative 4 right over here this is 0 that first integral is going to give us is going to give us this area right over here what's the area under the curve negative x minus 2 but under that curve or under that line and above the x-axis and the second integral is going to give us this area right over here between X plus 2 and and the x-axis going from negative 2 to 0 and so let's evaluate each of these and you might even be able just evaluate these with a little bit of triangle areas but let's just do this analytically or algebraically and so what's the antiderivative of negative x well that's negative x squared over 2 and then we have the minus the negative 2 or so this going to the antiderivative negative 2x we're going to evaluate that at at negative at negative two and negative four and so that part is going to be what negative two squared so it's the negative of negative two squared so it's negative 4 over 2 minus 2 times negative 2 so plus 4 so that's it evaluated at negative 2 and then minus if we evaluated it at negative 4 so minus so we're going to minus negative 4 squared is 16 over 2 minus 2 times negative 4 so that is plus plus 8 so what is that what is that going to give us so this is this is negative 2 this right over here is negative 8 so the second term right over here is just going to be equal to 0 did I do that right yeah the 16 over 2 it's negative and it's ok so this is just going to be 0 and this is negative 2 plus 4 which is going to be equal to 2 so what we have here in magenta is equal to 2 and what we have here in the blue well let's see this is the antiderivative of x squared over 2 plus 2x then evaluate it at 0 and negative 2 you evaluate this thing at 0 it's just going to be 0 and from that you're going to subtract negative 2 squared over 2 that is positive 4 over 2 which is positive 2 and then plus 2 times negative 2 so minus minus 4 and so this is going to be this is going to be the negative of negative 2 or positive 2 so it's 2 plus 2 and that makes sense that what we have in magenta here is 2 and what we have over here is 2 there's a symmetry here there is a symmetry here and so you add them all together and you get our integral is going to be equal to 4 and once again just as a reality check you could say look the height here is 2 the width the base here is 2 2 times 2 times 1/2 is indeed equal to 2 same thing over here so that's the more geometric argument for why that area is 2 that area is to add them together you get positive 4

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